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Difference between revisions of "1960 IMO Problems/Problem 6"

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==Solution==
 
==Solution==
 
{{solution}}
 
{{solution}}
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Part (a):
  
 
Let <math>R</math> denote the radius of the cone, and let <math>r</math> denote the radius of the cylinder and sphere. Let <math>l</math> denote the slant height of the cone, and let <math>h</math> denote the height of the cone.
 
Let <math>R</math> denote the radius of the cone, and let <math>r</math> denote the radius of the cylinder and sphere. Let <math>l</math> denote the slant height of the cone, and let <math>h</math> denote the height of the cone.
Line 21: Line 23:
 
<cmath>V_1 = \frac{1}{3}\pi R^2 h</cmath>
 
<cmath>V_1 = \frac{1}{3}\pi R^2 h</cmath>
 
<cmath>V_2 = \pi r^2 \times 2r = 2\pi r^3 = \frac{2R^3 h^3}{(l+R)^3}</cmath>
 
<cmath>V_2 = \pi r^2 \times 2r = 2\pi r^3 = \frac{2R^3 h^3}{(l+R)^3}</cmath>
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Now, we will compute the quantity <math>\frac{3}{\pi R^3} (V_1 - V_2)</math> and prove that it is always greater than <math>0</math>.
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Now, we will compute the quantity <math>\frac{3(l+R)^3}{\pi R^5 h} (V_1 - V_2)</math> and prove that it is always greater than <math>0</math>.
−
Let <math>x = \frac{h}{R}</math>. Clearly, <math>x</math> can be any positive real number.
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Let <math>x = \frac{h}{R}</math>. Clearly, <math>x</math> can be any positive real number. Define <math>W_1 = \frac{3(l+R)^3}{\pi R^5 h} V_1</math> and <math>W_2 = \frac{3(l+R)^3}{\pi R^5 h} V_2</math>. We will calculate <math>W_1</math> and <math>W_2</math> in terms of <math>x</math> and then compute the desired quantity <math>W_1 - W_2</math>.
  
 
It is easy to see that:
 
It is easy to see that:
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<cmath>V_1 = \frac{\pi R^3}{3} (\sqrt{x^2+1} + 1)^3</cmath>
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<cmath>W_1 = (\sqrt{x^2+1} + 1)^3</cmath>
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<cmath>V_2 = \frac{\pi R^3}{3} 6x^2</cmath>
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<cmath>W_2 = 6x^2</cmath>
  
 
Now, let <math>u = \sqrt{x^2+1}</math>. Since <math>x > 0</math>, it follows that <math>u > 1</math>. We now have:
 
Now, let <math>u = \sqrt{x^2+1}</math>. Since <math>x > 0</math>, it follows that <math>u > 1</math>. We now have:
−
<cmath>V_1 = \frac{\pi R^3}{3} (u + 1)^3</cmath>
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<cmath>W_1 = (u + 1)^3</cmath>
−
<cmath>V_2 = \frac{\pi R^3}{3} 6(u^2 - 1)</cmath>
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<cmath>W_2 = 6(u^2 - 1)</cmath>
  
−
Define <math>f(u) = \frac{3}{\pi R^3} (V_1 - V_2)</math>. It follows that:
+
Define <math>f(u) = W_1 - W_2</math>. It follows that:
 
<cmath>f(u) = (u+1)^3 - 6(u^2 - 1)</cmath>
 
<cmath>f(u) = (u+1)^3 - 6(u^2 - 1)</cmath>
 
<cmath>f(u) = u^3 - 3u^2 + 3u + 7</cmath>
 
<cmath>f(u) = u^3 - 3u^2 + 3u + 7</cmath>
 
<cmath>f(u) = (u-1)^3 + 8 > 8 > 0</cmath>
 
<cmath>f(u) = (u-1)^3 + 8 > 8 > 0</cmath>
  
−
We see that <math>f(u) > 8</math> for all allowed values of <math>u</math>. Thus, <math>V_1 - V_2 > \frac{8\pi}{3} R^3</math>, meaning that <math>V_1 \ne V_2</math>.
+
We see that <math>f(u) > 8</math> for all allowed values of <math>u</math>. Thus, <math>V_1 - V_2 > \frac{8\pi R^5 h}{3(l+R)^3}</math>, meaning that <math>V_1 > V_2</math>. We have thus proved that <math>V_1 \ne V_2</math>, as desired.
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 +
Part (b):
 +
 
 +
From our earlier work in calculating the volumes <math>V_1</math> and <math>V_2</math>, we easily see that:
 +
<cmath>\frac{V_1}{V_2} = \frac{(u+1)^3}{6(u^2 - 1)}</cmath>
 +
Re-expressing and simplifying, we have:
 +
<cmath>\frac{6V_1}{V_2} = \frac{(u+1)^2}{u-1} = (u-1) + 4 + \frac{4}{u-1}</cmath>
 +
By the AM-GM Inequality, <math>(u-1) + \frac{4}{u-1} \ge 4</math>, meaning that <math>\frac{V_1}{V_2} \ge \frac{4}{3}</math>.
 +
Equality holds if and only if <math>u-1 = \frac{4}{u-1}</math>, meaning that <math>u=3</math> and <math>x = 2\sqrt{2}</math>.
 +
 
 +
If we check the case <math>x = \frac{h}{R} = 2\sqrt{2}</math>, we may calculate <math>V_1</math> and <math>V_2</math>:
 +
<cmath>V_1 = \frac{1}{3} \pi R^2 h = \frac{2\sqrt{2}}{3}\pi R^3</cmath>
 +
<cmath>V_2 = 2\pi r^3 = 2\pi \left(\frac{R\times 2\sqrt{2} R}{4R}\right)^3 = \frac{\pi}{\sqrt{2}} R^3</cmath>
 +
Indeed, we have <math>\frac{V_1}{V_2} = \frac{4}{3}</math>, meaning that our minimum of <math>k = \frac{4}{3}</math> can be achieved.
 +
 
 +
Thus, we have proved that the minimum value of <math>k</math> such that <math>V_1 = kV_2</math> is <math>\frac{4}{3}</math>.
 +
 
 +
Now, let <math>\theta</math> be the angle subtended by a diameter of the base of the cone at the vertex of the cone. We have the following:
 +
<cmath>\tan\left(\frac{\theta}{2}\right) = \frac{1}{x} = \frac{\sqrt{2}}{4}</cmath>
 +
From the double-angle formula for tangent,
 +
<cmath>\theta = \arctan\left(\frac{4\sqrt{2}}{7}\right)</cmath>
 +
This angle is easy to construct. Simply take any segment and treat it as a unit segment. Create a right triangle with legs of lengths <math>4\sqrt{2}</math> and <math>7</math>. This is straightforward, and the angle opposite the leg of length <math>4\sqrt{2}</math> will be the desired angle <math>\theta</math>.
 +
 
 +
It follows that we have successfully constructed the desired angle <math>\theta</math>.
  
 
==See Also==
 
==See Also==
  
 
{{IMO7 box|year=1960|num-b=5|num-a=7}}
 
{{IMO7 box|year=1960|num-b=5|num-a=7}}

Revision as of 00:22, 7 July 2015

Problem

Consider a cone of revolution with an inscribed sphere tangent to the base of the cone. A cylinder is circumscribed about this sphere so that one of its bases lies in the base of the cone. Let $V_1$ be the volume of the cone and $V_2$ be the volume of the cylinder.

a) Prove that $V_1 \neq V_2$;

b) Find the smallest number $k$ for which $V_1 = kV_2$; for this case, construct the angle subtended by a diamter of the base of the cone at the vertex of the cone.

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

Part (a):

Let $R$ denote the radius of the cone, and let $r$ denote the radius of the cylinder and sphere. Let $l$ denote the slant height of the cone, and let $h$ denote the height of the cone.

Consider a plane that contains the axis of the cone. This plane will slice the cone and sphere into a circle $\omega$ inscribed in an isosceles triangle $T$.

The area of $T$ may be computed in two different ways: \[[T] = \frac{1}{2} \times r \times (2l + 2R) = r(l+R)\] \[[T] = \frac{1}{2} \times 2R \times h = Rh\] From this, we deduce that $r = \frac{Rh}{l+R}$.

Now, we calculate our volumes: \[V_1 = \frac{1}{3}\pi R^2 h\] \[V_2 = \pi r^2 \times 2r = 2\pi r^3 = \frac{2R^3 h^3}{(l+R)^3}\] Now, we will compute the quantity $\frac{3(l+R)^3}{\pi R^5 h} (V_1 - V_2)$ and prove that it is always greater than $0$. Let $x = \frac{h}{R}$. Clearly, $x$ can be any positive real number. Define $W_1 = \frac{3(l+R)^3}{\pi R^5 h} V_1$ and $W_2 = \frac{3(l+R)^3}{\pi R^5 h} V_2$. We will calculate $W_1$ and $W_2$ in terms of $x$ and then compute the desired quantity $W_1 - W_2$.

It is easy to see that: \[W_1 = (\sqrt{x^2+1} + 1)^3\] \[W_2 = 6x^2\]

Now, let $u = \sqrt{x^2+1}$. Since $x > 0$, it follows that $u > 1$. We now have: \[W_1 = (u + 1)^3\] \[W_2 = 6(u^2 - 1)\]

Define $f(u) = W_1 - W_2$. It follows that: \[f(u) = (u+1)^3 - 6(u^2 - 1)\] \[f(u) = u^3 - 3u^2 + 3u + 7\] \[f(u) = (u-1)^3 + 8 > 8 > 0\]

We see that $f(u) > 8$ for all allowed values of $u$. Thus, $V_1 - V_2 > \frac{8\pi R^5 h}{3(l+R)^3}$, meaning that $V_1 > V_2$. We have thus proved that $V_1 \ne V_2$, as desired.

Part (b):

From our earlier work in calculating the volumes $V_1$ and $V_2$, we easily see that: \[\frac{V_1}{V_2} = \frac{(u+1)^3}{6(u^2 - 1)}\] Re-expressing and simplifying, we have: \[\frac{6V_1}{V_2} = \frac{(u+1)^2}{u-1} = (u-1) + 4 + \frac{4}{u-1}\] By the AM-GM Inequality, $(u-1) + \frac{4}{u-1} \ge 4$, meaning that $\frac{V_1}{V_2} \ge \frac{4}{3}$. Equality holds if and only if $u-1 = \frac{4}{u-1}$, meaning that $u=3$ and $x = 2\sqrt{2}$.

If we check the case $x = \frac{h}{R} = 2\sqrt{2}$, we may calculate $V_1$ and $V_2$: \[V_1 = \frac{1}{3} \pi R^2 h = \frac{2\sqrt{2}}{3}\pi R^3\] \[V_2 = 2\pi r^3 = 2\pi \left(\frac{R\times 2\sqrt{2} R}{4R}\right)^3 = \frac{\pi}{\sqrt{2}} R^3\] Indeed, we have $\frac{V_1}{V_2} = \frac{4}{3}$, meaning that our minimum of $k = \frac{4}{3}$ can be achieved.

Thus, we have proved that the minimum value of $k$ such that $V_1 = kV_2$ is $\frac{4}{3}$.

Now, let $\theta$ be the angle subtended by a diameter of the base of the cone at the vertex of the cone. We have the following: \[\tan\left(\frac{\theta}{2}\right) = \frac{1}{x} = \frac{\sqrt{2}}{4}\] From the double-angle formula for tangent, \[\theta = \arctan\left(\frac{4\sqrt{2}}{7}\right)\] This angle is easy to construct. Simply take any segment and treat it as a unit segment. Create a right triangle with legs of lengths $4\sqrt{2}$ and $7$. This is straightforward, and the angle opposite the leg of length $4\sqrt{2}$ will be the desired angle $\theta$.

It follows that we have successfully constructed the desired angle $\theta$.

See Also

1960 IMO (Problems)
Preceded by
Problem 5 		1 • 2 • 3 • 4 • 5 • 6 • 7 Followed by
Problem 7
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